Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, a[102][102], b[102][102], c[102][102];
bool check;
vector<pair<int, int> > v, v2;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) scanf("%d", &b[i][j]);
  for (int l = 0; l < 502; l++) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (a[i][j] < b[i][j]) {
          check = true;
          for (int k = 0; k < n; k++) {
            if (a[k][j] + 1 > b[k][j]) {
              check = false;
              break;
            }
          }
          if (check) {
            for (int k = 0; k < n; k++) a[k][j]++;
            v.push_back(make_pair(1, j + 1));
          } else {
            check = true;
            for (int k = 0; k < m; k++) {
              if (a[i][k] + 1 > b[i][k]) {
                check = false;
                break;
              }
            }
            if (check) {
              for (int k = 0; k < m; k++) a[i][k]++;
              v.push_back(make_pair(0, i + 1));
            } else {
              cout << -1 << endl;
              return 0;
            }
          }
        }
      }
    }
  }
  for (int l = 0; l < 502; l++) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (c[i][j] < b[i][j]) {
          check = true;
          for (int k = 0; k < m; k++) {
            if (c[i][k] + 1 > b[i][k]) {
              check = false;
              break;
            }
          }
          if (check) {
            for (int k = 0; k < m; k++) c[i][k]++;
            v2.push_back(make_pair(0, i + 1));
          } else {
            check = true;
            for (int k = 0; k < n; k++) {
              if (c[k][j] + 1 > b[k][j]) {
                check = false;
                break;
              }
            }
            if (check) {
              for (int k = 0; k < n; k++) c[k][j]++;
              v2.push_back(make_pair(1, j + 1));
            } else {
              cout << -1 << endl;
              return 0;
            }
          }
        }
      }
    }
  }
  if (v.size() > v2.size()) v = v2;
  printf("%d\n", (int)v.size());
  for (int i = 0; i < v.size(); i++) {
    if (v[i].first == 0)
      printf("row ");
    else
      printf("col ");
    printf("%d\n", v[i].second);
  }
  return 0;
}